如何解决这种 CodingBat java 方法之一？

Question

这是我的作业：

如果两个字符串不为空并且它们的第一个字符相同，我们会说它们“匹配”。循环然后返回给定的非空字符串数组，如下所示：如果字符串与数组中较早的字符串匹配，则交换数组中的 2 个字符串。当数组中的一个位置被交换时，它不再匹配任何东西。使用地图，只需通过数组一次即可解决此问题。

allSwap(["ab", "ac"]) → ["ac", "ab"]

allSwap(["ax", "bx", "cx", "cy", "by", "ay", "aaa", "azz"]) → ["ay", "by", "cy", "cx", "bx", "ax", "azz", "aaa"]

allSwap(["ax", "bx", "ay", "by", "ai", "aj", "bx", "by"]) → ["ay", "by", "ax", "bx", "aj", "ai", "by", "bx"]

Answer 1

在map中，存储第一个字母作为key，key的最近索引作为value。当地图中不存在字母时，将其添加到地图中。当地图中已经存在一个字母时，将其从地图中移除并与该索引交换。

/**
 * Swaps strings in the array that have the same first letter,
 * reading left to right. Once a string has been swapped, 
 * it will not be swapped again. The input array will be mutated.
 * 
 * @param  strings the strings to perform swaps from
 * @return         the strings after swapping
 */
public static String[] allSwap(final String[] strings) {
    // map of first characters, and the index where they were last seen
    final Map<Character, Integer> potentialSwap = new HashMap<>();

    for (int thisIndex = 0; thisIndex < strings.length; thisIndex++) {
        if (strings[thisIndex].isEmpty()) {
            continue; // skip empty strings
        }

        final Character firstChar = strings[thisIndex].charAt(0); // box charAt(0)
        // remove firstChar from the map. If it's not found, returns null
        final Integer potentialIndex = potentialSwap.remove(firstChar);

        if (potentialIndex != null) {
            final int thatIndex = potentialIndex; // unbox potentialIndex
            // swap values at thisIndex and thatIndex
            final String temp = strings[thatIndex];
            strings[thatIndex] = strings[thisIndex];
            strings[thisIndex] = temp;
        } else {
            // save the index for possible swapping later
            potentialSwap.put(firstChar, thisIndex); // box thisIndex
        }
    }

    return strings;
}

Ideone Demo

Answer 2

public String[] allSwap(String[] strings) {

    // map of first characters, and the index where they were last seen
     Map<Character, Integer> map = new HashMap();

    for (int i = 0; i < strings.length; i++) {


         char firstChar = strings[i].charAt(0); 
         // box charAt(0)
        // remove firstChar from the map. If it's not found, returns null
         Integer removedIndex = map.remove(firstChar);

        if (removedIndex != null) {
             int j = removedIndex; 
            // unbox potentialIndex
            // swap values at thisIndex and thatIndex
             String temp = strings[j];
            strings[j] = strings[i];
            strings[i] = temp;
        } else {
            // save the index for possible swapping later
            map.put(firstChar, i); 
        }
    }

    return strings;
}

Answer 3

这也可以！

 public String[] allSwap(String[] strings) {
      
      
      Map<String,Integer> map = new HashMap<String,Integer>();
      
      
      for(int i=0;i<strings.length;i++){
        
        if(!map.containsKey(strings[i].substring(0,1))){
          map.put(strings[i].substring(0,1),i);
        }
        else{
          String temp = strings[map.get(strings[i].substring(0,1))];
          strings[map.get(strings[i].substring(0,1))] = strings[i];
          strings[i] = temp;
          map.remove(strings[i].substring(0,1));
        }
        
      }
      
    return strings;
    }